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natta225 [31]
3 years ago
13

Ancient gold coins have been found in Greece and Italy.What property allow for gold to be stamped into coins ?

Physics
1 answer:
Veronika [31]3 years ago
8 0

Answer: Malleability

Explanation:

Gold in addition to being so rare, is the most malleable element there is. A Malleable element can be shaped and reshaped without it breaking and gold is so malleable that you can hit it till it becomes so flat, its transparent.

This malleability allowed for the ancient peoples to be able to shape and stamp gold into coins and their rarity made them quite valuable. Even today, investors still shape gold into exquisite designs of coins so it not only has monetary value, but aesthetic value as well.

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What determines the number of possible sublevels?
Talja [164]
<h2>Hello!</h2>

The answer is A.   the principal energy level

<h2>Why?</h2>

The principal energy level or principal quantum number n, tell us about the position of an determined electron in the energy levels relative to the greater average distance of an electron from the nucleus, the larger the value of n and the higher its energy.

The principal energy levels contains n sublevels, n^{2} orbitals and 2n^{2} electrons.

For example, the number 1, has one orbital which is contained in a energy sublevel, its called s orbital, and it's just an orbital with 2 electrons.

Have a nice day!

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3 years ago
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Charra [1.4K]

They're not hiding anywhere. They're all right there in the light that's shining into the mist.

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3 years ago
A runner starts from rest, runs for 30
aleksklad [387]

Answer:

u=0

t=30s

a=2m/s²

v=?

v= u+at

= 0+(30)(2)

= 60m/s

6 0
3 years ago
Hãy nêu sự tương tự giữa công của lực điện trong trường hợp này với công của trọng lực.​
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Answer:

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6 0
3 years ago
I'm completely stumped as to how to do this.
worty [1.4K]

Explanation:

You have already determined the components of the known forces so I won't repeat your work here. Since the resultant force \vec{\textbf{R}} and F1 are completely along the x-axis, we can conclude that

F_{2y} = F_{3y} \Rightarrow F_{3y} = F_3\cos{\theta} = 250\:\text{lb}

We can now solve for the magnitude of F_3:

F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(467)^2 + (250)^2}

\:\:\:\:=529.7\:\text{lb}

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\tan{\theta} = \dfrac{F_{3y}}{F_{3x}} = \dfrac{250}{467}

or

\theta = 49.2°

6 0
3 years ago
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